Talk:Order (group theory)

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Are there any WP articles currently which explain why any element of a finite group has finite order? (I don't need it explained to me; I'm just concerned it doesn't seem to be discussed anywhere, and is not immediately obvious to a group theory newcomer.) Dmharvey Talk 5 July 2005 18:14 (UTC)

Isn't that a finite group is defined as a group whose order is finite? -- Taku 11:00, 5 October 2005 (UTC)

No, a finite group is one which has a finite number of elements. It follows that every element has finite order, so it may mean the same thing, but it's not, strictly speaking, defined as such. J•A•K 13:19, 9 December 2005 (UTC)

Erm, the order of a group is the number of elements. Did Taku mean to ask whether a finite group was defined as a group all of whose elements are finite? Shawn M. O'Hare 23:20, 13 January 2006 (UTC)

1 Moving to finite group?
2 Clarification, part 1
3 Clarification, part 2
4 ???

[edit] Moving to finite group?

Am I the only one who feels strange about the fact that many well-known facts about finite groups (like how a group order is divided) are put here instead of finite group? Since when the order is infinite, little can be said so why don't we move the stuff to finite group and redirect this to group (mathematics)? -- Taku 11:19, 5 October 2005 (UTC)

Yes

[edit] Clarification, part 1

This might help clarify:

If a group has finite order, there are a finite number of elements in that group by definition.

For an element a in group G with identity e, consider the set {m is a natural number: a^m = e}. If this set is not empty, then by the Well-Ordering Principle, it has a least member. This least member is defined as the order of a.

[edit] Clarification, part 2

If G is a finite group of order n, and d is a divisor of n, then the number of elements in G of order d is a multiple of φ(d).

That should be:

If G is a finite group of order n, and d is a divisor of n, then the number of elements in G of order d is φ(d).

[edit] ???

"If every element in G is the same as its inverse (i.e., g = g-1), then ord(g) = 2 and consequently G is abelian since ab = (bb)ab(aa) = b(ba)(ba)a = ba."

Well...what about $Z_2 \times Z_2$ ? --VKokielov 02:21, 8 July 2006 (UTC)